Calculators don't really do algebra when solving equations. They perform calculations very quickly. The Newton Raphson method is a way to get within a certain error window of the solution.
We are trying to find the zeros of the polynomial f(x) = 2x4 + 3x3 - 4x2 - 3x + 2 using the Newton Raphson method which is an iterative method where x1 could be just about anything and xn+1 = xn - [f(xn)/f'(xn)] should zoom in on the nearest zero of the polynomial to our chosen x1.
First, let's take the derivative so we have it to work with. f'(x)=8x3 + 9x2 - 8x -3 (using power rule).
Note that the error between iterations (f(xn)/f'(xn)) is 0 if f(xn)=0, so we don't need to use the Newton Raphson method if we guess correctly right off the bat. Now, just looking at the coefficients, x=1 is one of the zeros (if you plug in 1, you're just adding the coefficients). If you evaluate f(1), you'll find that x=1 is in fact one of our zeros. So is x=-1 (hearkening back to some algebra methods for polynomials, you can flip the sign on coefficients in front of odd exponent terms and add the new exponents). The third "nice" value to guess with is x=0, but f(0) is not 0.
Let's try x1=0 to seed our Newton Raphson iterations because it's a nice central place to start our guess work and it's simple to calculate f(0) and f'(0) without a calculator. Since we know that x=-1 and x=1 are solutions to f(x)=0, this will find out if there is a solution in between -1 and 1.
x1 = x0 - [f(x0)/f'(x0)] = 0 - [2/-3] = 2/3.
x2 = 2/3 - [f(2/3)/f'(2/3)] (by substituting into the definition of Newton Raphson iteration)
= 2/3 - [(-40/81)/(-53/27)] (by some hand calculations that should be double checked)
= 2/3 - 40/159
= 22/53 (it's not getting pretty, but it is getting pretty close to 1/2).
x3 = 22/53 - [f(22/53)/f'(22/53)]
= 22/53 - [(2678400/7890481)/(-624963/148877)]
Which approximately equals 0.496. It's safe to say we're getting close to 1/2. We can evaluate f(1/2) to double check. Yep, it works.
Thus we have found 3 real rational solutions to a 4th degree polynomial. Both complex and irrational roots come in pairs, so the 4th and final root will be a real rational number. If you're suspicious that there might be another root between -1 and 1, you could seed your iteration with -1/2 to check between -1 and 0, 1/4 to check between 0 and 1/2, or 3/4 to check between 1/2 and 1. If you remember Descartes' Rule of Signs, you can check how many positive and negative rational roots f(x) could have. f(x) has 2 sign changes, so it can have at most 2 positive rational roots. By elimination, the 4th root must be negative. Or you can use Descartes' to note that f(-x) has 2 sign changes as well, so f(x) must have either 2 or 0 negative rational roots (which confirms that the last root is negative). All of that just means don't bother testing positive x0 values. Try x0=-3, and take the iterations until you see where they're headed.
For graphing, pick one of the iterations (either the one that lead to 1/2 or the one I left to you to find the last solution). When you draw your graph, you'll feel pretty dorky because you'll know what the function f(x) looks like near that value. Draw f(x) (and use a fairly zoomed in window of values). Draw tangent lines through (xn, f(xn)) with slope f'(xn) for each iteration n, and make sure you bring them through the x-axis. Note that the tangent line for the next iteration starts directly above or below the previous x-intercept, and the x-intercepts of the tangent lines get closer to the desired value.
Your instructor may be extra happy with you if you draw the graph for both sets of iterations. There is no need to draw a graph if you didn't have to perform iterations to find the solution, so at most you should be drawing two full graphs.
The Newton Raphson method is tedious on written homework. It is valuable to understand how computers "solve" certain problems by honing in on a good approximation by quickly calculating iterations, and that the approximation can be improved by performing more iterations. Computer programmers want computers to provide accurate results but not bog down by performing too many iterations, so most computers are set to keep iterating until the error is within a set window. At some point, you may have used the abilities of a graphing calculator to find the zero of a function, and it may have produced slightly different results when your viewing window was different. The calculator was guessing how accurate you wanted the result.